Question

The angle of depression of the top of a tree as observed from the roof of a house is 30 ft high is found to be 30 degree if the distance between the house and tree is 10ft hind height of tree

Answer 1

we have taken height of tree DC and height of house AB and distance between tree and house BC and then calculated BE. BE is equal to DC because both house and tree are perpendicular to the ground

Answer 2

The height of the tree is 24.23 ft.

Given:

• The angle of depression on the top of a tree as observed from the roof of a house is 30 ft high and is found to be 30 degree.
• If the distance between the house and tree is 10 ft.

To find:

• Find the height of the tree.

Solution:

Concept to be used:

• Apply trigonometric ratio.

Step 1:

Write the values.

Let the Height of tree be 'x' ft.

*See the attachment (Not per measurement).

DC is tree, and

$\bf DC=BH = x$

So,

$\bf AB = (30 - x) \: ft$

$\bf BC= HD = 10 \: ft$

Step 2:

Find the height of the tree.

Apply trigonometric ratio in ∆ABC.

$tan \: \theta=\frac{AB}{BC}$

$tan \: {30}^{ \circ} =\frac{30 - x}{10}$

$\frac{1}{ \sqrt{3} } = \frac{30 - x}{10}$

$\frac{10}{ \sqrt{3} } = (30 - x)$

$\frac{10}{ \sqrt{3} } - 30 = - x$

$x = 30 - \frac{10}{ \sqrt{3} }$

$x = \frac{30 \sqrt{3} - 10 }{ \sqrt{3} }$

Multiply and divide by √3.

$x = \frac{(30 \sqrt{3} - 10) \sqrt{3} }{3}$

$x = \frac{90 - 10 \sqrt{3} }{3} \: ft$

For simplify further, put $\sqrt{3} = 1.732$

So,

$x = \frac{90 -17.32 }{3}$

$\bf DC = 24.23 \: ft$

Thus,

The height of the tree is 24.23 ft or ≈24 ft.

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